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# EDO 9th Boyce Cap9 Resolução

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is at (3 , 2). The system is solved numerically, with
y(0) = 2 and x(0) = 3.5 , 4.0 , 4.5 , 5.0 . The resulting periods are shown in the table:
x(0) = 3.5 x(0) = 4.0 x(0) = 4.5 x(0) = 5.0
T 7.26 7.29 7.34 7.42
The actual amplitude steadily increases as the amplitude increases.
9. The system
dx
dt
= a x(1\u2212 y
2
)
dy
dt
= b y(\u22121 + x
3
)
is solved numerically for various values of the parameters. The initial conditions
are x(0) = 5 , y(0) = 2 .
500 Chapter 9. Nonlinear Differential Equations and Stability
(a) a = 1 and b = 1 :
The period is estimated by observing when the trajectory becomes a closed curve.
In this case, T \u2248 6.45 .
(b) a = 3 and a = 1/3 , with b = 1 :
For a = 3 , T \u2248 3.69 . For a = 1/3 , T \u2248 11.44 .
(c) b = 3 and b = 1/3 , with a = 1 :
For b = 3 , T \u2248 3.82 . For b = 1/3 , T \u2248 11.06 .
(d) It appears that if one of the parameters is fixed, the period varies inversely with
9.5 501
the other parameter. Hence one might postulate the relation
T =
k
f(a , b)
.
10.(a) Since T = 2pi/
\u221a
ac , we first note that\u222b A+T
A
cos(
\u221a
ac t+ \u3c6)dt =
\u222b A+T
A
sin(
\u221a
ac t+ \u3c6)dt = 0 .
Hence
x =
1
T
\u222b A+T
A
c
\u3b3
dt =
c
\u3b3
and y =
1
T
\u222b A+T
A
a
\u3b1
dt =
a
\u3b1
.
(b) One way to estimate the mean values is to find a horizontal line such that
the area above the line is approximately equal to the area under the line. From
Figure 9.5.3, it appears that x \u2248 3.25 and y \u2248 2.0 . In Example 1 , a = 1 , c = 0.75 ,
\u3b1 = 0.5 and \u3b3 = 0.25 . Using the result in part (a), x = 3 and y = 2 .
(c) The system
dx
dt
= x(1\u2212 y
2
)
dy
dt
= y(\u22123
4
+
x
4
)
is solved numerically for various initial conditions.
x(0) = 3 and y(0) = 2.5 :
502 Chapter 9. Nonlinear Differential Equations and Stability
x(0) = 3 and y(0) = 3.0 :
x(0) = 3 and y(0) = 3.5 :
x(0) = 3 and y(0) = 4.0 :
It is evident that the mean values increase as the amplitude increases. That is, the
mean values increase as the initial conditions move farther from the critical point.
9.5 503
12.(a) The equilibrium points are the solutions of the system
x(a\u2212 \u3c3x\u2212 \u3b1y) = 0
y(\u2212c+ \u3b3x) = 0.
If x = 0, then y = 0. If y = 0, then x = a/\u3c3. The third solution is found by
substituting x = c/\u3b3 into the first equation. This implies that y = a/\u3b1\u2212 \u3c3c/(\u3b3\u3b1).
So the equilibrium solutions are (0, 0), ( a\u3c3 , 0) and (
c
\u3b3 ,
a
\u3b1 \u2212 \u3c3c\u3b3\u3b1 ). When \u3c3 is increasing,
the critical point ( a\u3c3 , 0) moves to the left and the critical point (
c
\u3b3 ,
a
\u3b1 \u2212 \u3c3c\u3b3\u3b1 ) moves
down. The assumption a > \u3c3c\u3b3 is necessary for the third equilibrium to be in the
first quadrant. (When a = \u3c3c\u3b3 , then the two nonzero critical points coincide.)
(b) The Jacobian of the system is
J =
(
a\u2212 2\u3c3x\u2212 \u3b1y \u2212\u3b1x
\u3b3y \u2212c+ \u3b3x
)
.
This implies that at the origin
J(0, 0) =
(
a 0
0 \u2212c
)
,
which implies that the origin is a saddle point. (a > 0 and c > 0 by our assumption.)
At the critical point ( a\u3c3 , 0)
J(
a
\u3c3
, 0) =
(\u2212a \u2212\u3b1a/\u3c3
0 \u2212c+ \u3b3a/\u3c3
)
,
which implies that this critical point is also a saddle as long as our assumption
a > \u3c3c\u3b3 is valid.
At the critical point ( c\u3b3 ,
a
\u3b1 \u2212 \u3c3c\u3b3\u3b1 )
J(
c
\u3b3
,
a
\u3b1
\u2212 \u3c3c
\u3b3\u3b1
) =
( \u2212\u3c3c/\u3b3 \u2212\u3b1c/\u3b3
\u3b3a/\u3b1\u2212 \u3c3c/\u3b1 0
)
,
which implies that this equilibrium point is either an asymptotically stable spiral
or node.
13.(a) The equilibrium points are the solutions of the system
x(1\u2212 x
5
\u2212 2y
x+ 6
) = 0
y(\u22121
4
+
x
x+ 6
) = 0.
If x = 0, then y = 0. If y = 0, then x = 5. The third equilibrium point can be
found by setting 1/4 = x/(x+ 6), which gives x = 2 and then y = 2.4. So the
critical points are (0, 0), (5, 0) and (2, 2.4).
(b) The Jacobian of the system is
J =
(
1\u2212 2x5 \u2212 12y(x+6)2 \u2212 2xx+6
6y
(x+6)2 \u2212 14 + xx+6
)
.
504 Chapter 9. Nonlinear Differential Equations and Stability
This implies that at the origin
J(0, 0) =
(
1 0
0 \u22121/4
)
,
which implies that the origin is a saddle point.
At the critical point (5, 0)
J(5, 0) =
(\u22121 \u221210/11
0 9/44
)
,
which implies that this critical point is also a saddle point.
At the critical point (2, 2.4)
J(2, 2.4) =
(\u22121/4 \u22121/2
9/40 0
)
,
which implies that this equilibrium point is an asymptotically stable spiral.
15.(a) Solving for the equilibrium of interest we obtain
x =
E2 + c
\u3b3
y =
a
\u3b1
\u2212 \u3c3
\u3b1
· E2 + c
\u3b3
\u2212 E1
\u3b1
.
So if E1 > 0 and E2 = 0, then we have the same amount of prey and fewer predators.
(b) If E1 = 0 and E2 > 0, then we have more prey and fewer predators.
(c) If E1 > 0 and E2 > 0, then we have more prey and even fewer predators.
16.(b) The equilibrium solutions are given by the solutions of the system
x(1\u2212 y
2
) = H1
y(\u22123
4
+
x
4
) = H2.
Now if H2 = 0, then x = 3 and the first equation gives y = 2\u2212 2H1/3. This means
we have the same amount of prey and fewer predators.
(c) If H1 = 0, then y = 2 and the second equation gives x = 3 + 2H2. This means
we have the same amount of predators and more prey.
9.6 505
(d) If H1 > 0 and H2 > 0, then the second equation gives (x\u2212 3)y = 4H2 and
using this we obtain from the first equation that x(1\u2212 2H2x\u22123 ) = H1. This gives the
quadratic equation x2 \u2212 (3 + 2H2 +H1)x+ 3H1 = 0. Now at the old value x = 3
this expression is \u22126H2, so there is a root which is bigger than x = 3. The other
root of the quadratic equation is closer to 0, so the equilibrium increases here: we
have more prey. (Check this with e.g. H1 = H2 = 1: the roots are 3±
\u221a
6, so both
of the original roots get bigger.) A similar analysis shows that we will have fewer
predators in this case.
9.6
2. We consider the function V (x, y) = a x2 + c y2 . The rate of change of V along
any trajectory is
V\u2d9 = Vx
dx
dt
+ Vy
dy
dt
= 2ax(\u22121
2
x3 + 2xy2) + 2cy(\u2212y3) =
= \u2212ax4 + 4ax2y2 \u2212 2cy4.
Let us complete the square now the following way:
V\u2d9 = \u2212ax4 + 4ax2y2 \u2212 2cy4 = \u2212a(x4 \u2212 4x2y2)\u2212 2cy4 =
= \u2212a(x2 \u2212 2y2)2 + 4ay4 \u2212 2cy4 = \u2212a(x2 \u2212 2y2)2 + (4a\u2212 2c)y4.
If a > 0 and c > 0 , then V (x, y) is positive definite. Clearly, if 4a\u2212 2c < 0, i.e.
when c > 2a, then V\u2d9 (x, y) is negative definite. One such example is V (x, y) =
x2 + 3 y2. It follows from Theorem 9.6.1 that the origin is an asymptotically stable
critical point.
4. Given V (x, y) = a x2 + c y2 , the rate of change of V along any trajectory is
V\u2d9 = Vx
dx
dt
+ Vy
dy
dt
= 2ax(x3 \u2212 y3) + 2cy(2xy2 + 4x2y + 2y3) =
= 2a x4 + (4c\u2212 2a)xy3 + 8c x2y2 + 4c y4.
Setting a = 2c ,
V\u2d9 = 4c x4 + 8c x2y2 + 4c y4 \u2265 4c x4 + 4c y4.
As long as a = 2c > 0 , the function V (x, y) is positive definite and V\u2d9 (x, y) is also
positive definite. It follows from Theorem 9.6.2 that (0 , 0) is an unstable critical
point.
5. Given V (x, y) = c(x2 + y2) , the rate of change of V along any trajectory is
V\u2d9 = Vx
dx
dt
+ Vy
dy
dt
= 2c x [y \u2212 xf(x, y)] + 2cy [\u2212x\u2212 yf(x, y)] =
= \u22122c(x2 + y2)f(x, y) .
506 Chapter 9. Nonlinear Differential Equations and Stability
If c > 0 , then V (x, y) is positive definite. Furthermore, if f(x, y) is positive in some
neighborhood of the origin, then V\u2d9 (x, y) is negative definite. Theorem 9.6.1 asserts
that the origin is an asymptotically stable critical point. On the other hand, if
f(x, y) is negative in some neighborhood of the origin, then V (x, y) and V\u2d9 (x, y) are
both positive definite. It follows from Theorem 9.6.2 that the origin is an unstable
critical point.
9.(a) Letting x = u and y = u \u2032, we obtain the system of equations
dx
dt
= y
dy
dt
= \u2212g(x)\u2212 y .
Since g(0) = 0 , it is evident that (0 , 0) is a critical point of the system. Consider
the function
V (x, y) =
1
2
y2 +
\u222b x
0
g(s)ds .
It is clear that V (0, 0) = 0 . Since g(u) is an odd function in a neighborhood of
u = 0 , \u222b x
0
g(s)ds > 0 for x > 0 ,
and \u222b x
0
g(s)ds = \u2212
\u222b